I came across OEIS sequences A190020 and A190019, and I noticed that they seemed to grow at a similar rate.
A190020: Number of obtuse triangles on a (n X n)-grid (or geoboard)
A190019: Number of acute triangles on a (n X n)-grid (or geoboard)
In particular, I'm interested in their ratio, which gives a sense of the relative frequency of obtuse vs acute triangles when points are chosen in a square.
Empirically, it looks like $\displaystyle \lim_{n \rightarrow \infty}\frac{A190020(n)}{A190019(n)} \approx 2.6391$. (See graph of the ratio below.)
- What is a heuristic for why the limit appears to be finite and greater than 1?
- How do you prove a finite upper bound for this ratio?
- How do you prove a nonzero lower bound for this ratio?
- Does this limit converge? Is it known what it converges to?
- If three points are selected uniformly at random to form a triangle $\Delta$ in the unit square $[0,1] \times [0,1]$, does $$\frac{P(\Delta \text{ is obtuse})}{P(\Delta \text{ is acute})} = \lim_{n \rightarrow \infty}\frac{A190020(n)}{A190019(n)}?$$
I have an outline for the calculation. It is a six-fold integral, for the six coordinates of the triangle. I think the six-fold integral can be evaluated to a single integral, which has several dozen elementary terms. I would need Maple or Mathematica to finish this.
Let $S$ be the unit square $[0,1]\times[0,1]$.
Let $A$ and $B$ have coordinates $A(x,y)$, and $B(x+r\cos\theta,y+r\sin\theta)$
Let the ray $AB$ meet the edge of $S$ at $F$. Let the line through $A$, perpendicular to $AB$, meet the square's edges at $D$ and $E$.
The probability $P$ that the triangle is obtuse is three times the probability that angle $\angle BAC$ is obtuse. In the following formula, $I_{BAC \text{ is obtuse}}$ is $1$ if the angle at $A$ is obtuse, and $0$ otherwise.
$$P = 3\iint_S da\iint_S db \iint_S dc\text{ } I_{BAC \text{ is obtuse}}$$
The integral over $c$ is the area bounded by $S$ and $DE$, on the opposite side from $B$. This has a dozen formulas, depending on which edges $D$ and $E$ are on. Let $T(x,y,\theta)$ be this area. The specific formula for the picture is $\frac12(x+y\tan\theta)(y+x\cot\theta)$
$$T(x,y,\theta)=\iint_S dc \text{ }I_{BAC\text{ is obtuse}}\\ P = 3\iint_Sda\iint_Sdb\text{ } T(x,y,\theta)$$
One can integrate over $b$ using polar coordinates centered at $A(x,y)$. Since $T$ is independent of $r$, this can be integrated out $$\iint_S db T(x,y,\theta)=\int dr \int d\theta \text{ }rT(x,y,\theta)\\ = \int d\theta\frac12(AF)^2 T(x,y,\theta)\\ P = 3\iint_S da \int_0^{2\pi}d\theta \frac12(AF)^2 T(x,y,\theta)$$ As $\theta$ runs through $2\pi$, there are twelve different formulas for $(AF)^2T(x,y,\theta)$, depending on which sides $D$, $E$ and $F$ are on. All twelve have an elementary integral. We can thus reach a formula $$U(x,y) = \int_0^{2\pi}d\theta\frac12(AF)^2 T(x,y,\theta)\\ P = 3\iint_S \text{ }dx\text{ } dy\text{ }U(x,y)$$
$U(x,y)$ has several dozen terms, involving $x^i(1-x)^jy^k(1-y)^l$ for exponents that run from $-2$ to $4$. It also has terms involving $\log x,\log (1-x),\log y$ and $\log(1-y)$. The log terms are multiplied by polynomials in $x$ and $y$; the exponents for those terms are not negative.
$U(x,y)$ has a different form if $A$ is in the yellow region, or orange region; and fourteen other formulas in the rest of the square. By symmetry, we just need to integrate over the yellow and orange regions, and multiply by eight because of the square's symmetry.
Each formula of $U(x,y)$ must be integrated over the appropriate region. Suppose $U_1(x,y)$ is the formula for the yellow region and $U_2$ for the orange region.
$$P = 24\int_0^{1/2}dx\int_{1/2-\sqrt{1/4-x^2}}^xdy \text{ }U_1(x,y) + \\ 24\int_0^{1/2}dx\int_0^{1/2-\sqrt{1/4-x^2}}dy\text{ } U_2(x,y) $$
I think the $y$ integrals have elementary form, but since the limits are on the circular arc, I haven't found an elementary form for the final $dx$ integral.